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Solutions to Homogeneous 2nd-order Linear D.E.s

  1. May 7, 2008 #1
    ,so in class we're covering second-order D.E.s...

    How exactly was it determined/derived that all solutions to homogeneous 2nd-order linear differential equations with constant coefficients (ay'' + by' + cy =0, where a,b,c are real constants) are of the form y=ert.

    Not asking for anyone to "do [my] homework" for me - I'm just curious. In lecture, and on the professor's website it simply says, "The reason for this is that long ago some geniuses figured this stuff out and it works." immediately after introducing the form of the solutions.

    ,also while I'm on the topic I have a question as to WHY we should continue considering the linear combination of solutions to these equations?

    ..ie so suppose we're given y''-y=0

    ,with a little work we can see that y1=2et and y2=5e-t are two possible solutions to the D.E.

    ,and it can be seen that c1y1+c2y2=y is also a solution

    ,why should/do we continue to only concentrate on the linear combination, when we could simply return to the multiplies of solutions y1 and y2 to solve any general or initial value problem?
  2. jcsd
  3. May 7, 2008 #2
    Well, not that i know much on differential equations, but here are my thoughts on how they came up with

    [tex] y=e^{rt}[/tex] as a solution of say a 2nd-order diff. eq with constant coefficients of the form


    I think that the first person who figured this out must have reasoned something like this:

    He should have noticed that every derivative of [tex]y=e^{rt}[/tex] is a r multiple of the previous one. that is

    [tex]y^{(n)}=r^ne^{rt}[/tex] , so he should have noticed that if we substitute this one in our equation then we get sth

    [tex] e^{rt}(ar^2+br+c)=0[/tex] so this means that if we can find a r such that

    [tex]ar^2+br+c=0[/tex] then we are done.
  4. May 7, 2008 #3
    Because for some initial conditions, you might get stupid dead ends if you don't consider the general solution. Suppose your solution is c1 cos(t) + c2 sin(t). If your initial conditions are such that you have 3 sin(t) as a solution and if you don't consider sin(t) part in the first place you would say this is nonsense but it is not. Probably that is why it is called a general solution.
  5. May 7, 2008 #4


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    You can reduce a 2nd order linear ODE to a system of 2 linear 1st order ODE. Or in general you can transform a nth order linear ODE to a system of n 1st order linear ODEs. An example on how to reduce a 3rd order linear ODE to a system of 3 1st order linear ODEs: https://www.physicsforums.com/showthread.php?t=232512

    As for why y = e^rt works as a general solution it's due to something called Picard's existence and uniqueness theorem. See: http://en.wikipedia.org/wiki/Picard's_existence_theorem

    Basically what this means (without going into the proof) is that once you have found a unique solution to a first order linear ODE, you have found all the possible solutions. Now, when you have a 2nd order linear ODE, you end up with two 1st order linear ODEs. Each of them would have their own unique solution, so that means you have 2 solutions. As for why we consider a linear combination instead of individual multiples of solutions y1 and y2, it's because of something known as the superposition principle: If both y1 and y2 satisfy a homogenous linear differential equation, then any linear combination of them would also satisfy it. Note that y1 and y2 themselves are simply special cases of c1y1 + c2y2; they may be obtained by setting c1=1,c2=0 and c1=0,c2=1. So in a sense, considering their linear combination is a stronger result because it encompasses the original solutions themselves.
  6. May 7, 2008 #5


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    They can't. There's more to it than that. I presume you will learn about other, more general solutions later in class.

    For one thing be sure you understand that you are talking about linear homogeneous differential equations with constant coefficients. Since there are no other functions involved by y and its derivatives, they had better be the same "kind" of function. If for example, you tried a y= log(x) as a solution, all of its derivatives are negative powers of x- you will never get anything to cancel the "log(x)" term itself. Since the derivative of an exponential function is an exponential function, it makes a perfect first "guess".

    Because multiples of solutions will not give all solutions. Neither Cet nor Ce-t will satisfy y(0)= 1, y'(0)= 2. Was that what you meant by "multiples of solutions"? You yourself write it can be seen that c1y1+c2y2=y is also a solution" and that is the general linear combination of the two functions.
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